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The revenue function is essentially a mathematical representation that helps us understand how much money a company can earn based on certain variables, such as the quantity of items sold. In this specific problem, the revenue function changes depending on the number of shoe pairs ordered.
When fewer than 50 pairs are ordered, each pair costs a fixed price of $40. Therefore, the revenue function for less than 50 pairs is simply calculated as \( R = 40n \), where \( n \) is the number of pairs. This is a straightforward linear equation.
However, when the order is 50 pairs or more, the price per pair is discounted by 4 cents for each pair ordered. This causes the revenue function to change to a quadratic form: \( R = n(40 - 0.04n) \), which simplifies to \( R = 40n - 0.04n^2 \).
This quadratic equation reflects the fact that as more are ordered, the price drops, influencing the total revenue.

Price per pair or unit pricing is a crucial element, as it determines the total revenue from sales. Price is influenced by the size of the order in this scenario.
If fewer than 50 pairs (\( n < 50 \)) are ordered, the price per pair remains constant at \(40. There's nothing complicated about this—it simply involves multiplying \)40 by the number of pairs.
However, the situation changes when the order size increases to 50 or more pairs. Here, instead of a fixed price, the price per pair decreases linearly by 4 cents with each additional pair ordered.
This forms the expression \( 40 - 0.04n \) for the price per pair when \( n \geq 50 \). As \( n \) grows, the total price reduction accumulates, impacting the overall revenue.

Critical points in calculus are where you can find the maximum or minimum values for a given function. They occur where the derivative of the function equals zero.
To find where the revenue is maximized for the company in this problem, take the derivative of the revenue function \( R = 40n - 0.04n^2 \). Calculating the derivative gives us \( \frac{dR}{dn} = 40 - 0.08n \).
By setting this derivative to zero, \( 40 - 0.08n = 0 \), we solve for \( n \), leading to the critical point of \( n = 500 \).
This critical point indicates the order size that maximizes the revenue within the range 50 to 600 pairs.
Checking the value at the boundaries and this critical point helps determine where the maximum revenue occurs.

Revenue maximization is the goal of determining the best order size or price that maximizes the company's earnings.
In this problem, to ensure maximum revenue, we need to evaluate the revenue at various key points: the calculated critical point and the endpoints of the order range.
For \( n = 500 \), the revenue can be computed as \( R = 40(500) - 0.04(500)^2 \). If this gives a higher revenue compared to the values at the endpoints, then 500 pairs is indeed the optimal order size.
Likewise, calculating revenue at the endpoints, such as \( n = 50 \) and \( n = 600 \), ensures that 500 is the point of maximum earnings.

• Evaluating \( R(50) = 40(50) - 0.04(50)^2 \)
• Evaluating \( R(600) = 40(600) - 0.04(600)^2 \)

Through this process, we validate that the maximum revenue is indeed achieved at \( n = 500 \). This verification is crucial to confirm the solution's correctness, ensuring that no greater revenue could be obtained within the given conditions.